THE INVERSE SQUARE LAW 21 1 



Now we can show by experiment that any other pole can be 

 placed at C instead of the unit pole and there will still be 

 equality of the two actions on it. This may easily be carried out 

 in a slightly modified form by bringing two vertical long bar 

 magnets up E and W of a given compass needle with their like 

 poles towards it and on the same level. If the magnets are so 



or 

 >n m,' I 



f <* -> 4 *' > 



ACs 



FIG. 100. 



adjusted that the compass remains in the meridian, any other 

 compass needle placed in the same position will also lie in the 

 meridian. 



If then we place a pole or strength m at C it acts on 1 at B 



with a force -y, 2 an( l is reacted on by an equal force which balances 



the action of m upon it. Hence the force exerted by won in = -y- 



/ . 1 _ in \ _ mm 



Y \ *<r jJ =i "3*"' 



We have already pointed out that we may assign the algebraic 

 + and to the two kinds of poles inasmuch as they have opposite 

 actions tending to neutralise each other when acting under similar 

 circumstances on the same pole. A NSP is always considered to 

 be + , a SSP to be . The force between two poles is therefore 



- and is a repulsion or an attraction according as their product 



is + or . 



Magnetic intensity. The force which would act on a 

 unit pole placed at any point in a magnetic field is the magnetic 

 intensity at that point. The unit intensity is termed one gauss. 

 We may conveniently denote the magnetic intensity by the letter 

 H. If a pole m be placed at a point where the intensity is H the 

 force acting upon it is mH.. 



Geometrical construction for the direction of the 

 intensity in the field of a bar magnet. On the supposition 

 that a magnet consists of two poles concentrated at two points a 

 given distance apart, the equations to the lines of force may easily 

 be determined. A construction for drawing the lines will be 

 found in the article Magnetism (Ency. Brit. 9th ed. p. 230). 

 The following construction will give the direction of the intensity 

 at any point. Let NS, Fig. 161, be the two poles, P a point in the 

 field, C the middle point of NS. From S, the nearer pole to P, 

 draw SO perpendicular to PS, meeting CO, the perpendicular to 



